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Given $\left\{A_n\right\}$ are i.i.d. Show that $E(|A_1|)< \infty$ $\iff \ P\left\{|A_n| > n \ \ \text{i.o}\right\} = 0$.

My attempt: ($\Rightarrow$) Assume $E(|A_1|)< \infty$. Since $\left\{A_n\right\}$ are i.i.d, by the Strong Law of Large Number, $\frac{1}{n}\sum_{i=1}^{n} |A_i|\rightarrow E(|A_1|)$ almost surely. This is equivalent to $\forall\ \epsilon > 0$, $\lim_{n\rightarrow \infty} P(\cup_{k\geq n}|(\frac{1}{k}\sum_{i=1}^{k} |A_i|)-E(|A_1|)|\geq \epsilon)= 0$. This implies for sufficiently large $k$, $\frac{1}{k}\sum_{i=1}^{k} |A_i| - E(|A_1|)\ <\ \epsilon$. Since we don't know whether $E(|A_1|) > 1$ or not, we cannot choose $\epsilon = 1-E(|A_1|)$ to get $|A_k|$ bounded above by $k$. Could someone please help with this part?

($\Leftarrow$) If $P\left\{|A_n| > n \ \ \text{i.o}\right\} = 0$, this implies for sufficiently large $k$, $k$ is fixed, $|A_k|< k$. This implies $E(|A_k|) = E(|A_1|) < E(k) = k < \infty$ (the first equality is due to $\left\{A_n\right\}$ are iid).

My question: Could someone help complete my "solution" above for the forward direction? If I'm on the wrong track, please help point that out to me.

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  • $\begingroup$ Well, this might seem silly, but can you tell me what i.o. means? $\endgroup$ – Qwerty Oct 7 '16 at 12:34
  • $\begingroup$ @Qwerty: i.o = infinitely often. And no, it's not silly. $\endgroup$ – user177196 Oct 8 '16 at 1:45
  • $\begingroup$ Well, can you be kind enough to specify the defnition of i.o.? $\endgroup$ – Qwerty Oct 8 '16 at 7:14
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The forward part is an easy consequence of Borel-Cantelli'e lemma. First of all see that: $$ \mathbb E(|A_1|)\geq \sum_{n=1}^\infty \mathbb P(|A_1|>n). $$ Using i.i.d. assumption: $$ \sum_{n=1}^\infty \mathbb P(|A_1|>n)= \sum_{n=1}^\infty \mathbb P(|A_n|>n). $$ So $\sum_{n=1}^\infty \mathbb P(|A_n|>n)<\infty$ and therefore from Borel-Cantelli's lemma: $$ \mathbb P(|A_n|>n , \rm{ i.o. })=0. $$

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  • $\begingroup$ thank you for your help. I don't think we have the coefficient $n$ in the first inequality? Otherwise, great proof (I should have thought of that:P). Btw, could you help verify if my solution for the reverse direction is correct? $\endgroup$ – user177196 Oct 4 '16 at 13:26
  • $\begingroup$ Your reverse direction seems fine to me. I also think that the coefficient $n$ should be there. That's why something stronger can be proved using bounded first moment. $\endgroup$ – Arash Oct 4 '16 at 13:28
  • $\begingroup$ The correct inequality is: for any $X\geq 0$, $\sum_{k=1}^{\infty} P(X > k)\leq E(X)\leq \sum_{k=0}^{\infty} P(X > k)$ $\endgroup$ – user177196 Oct 4 '16 at 13:29
  • $\begingroup$ Yes you are right. I mixed it up with something else. $\endgroup$ – Arash Oct 4 '16 at 13:35

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